A flutist assembles her flute in a room where the speed of sound is 342 m/sm/s . When she plays the note A, it is in perfect tune with a 440 Hz Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 346 m/sm/s .1.How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?
2.How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?

A) beats per second she will hear if she now plays the note A as the tuning fork is sounded = 5.13 beats per second

B) length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork = 0.0045m

Explanation:

A) First of all, wavelength = v/f

Where v is speed of wave and f is frequency.

Thus, wavelength of the sound wave of Note A is;

f2 = 440 Hz and v = 342m/s

λ = 342/440 = 0.7773m

Now, since the air inside the note was warmed after a while, the wave will will have a new frequency which we'll call (f1) and and new speed (v'), thus;

f2 = v'/λ = 346/0.7773 = 445.13 Hz

Now let's calculate beat frequency(fbeat).

fbeat = (f1 - f2)

So fbeat = 445.13 - 440 = 5.13Hz or 5.13 beats per second

B) Now, frequency of standing wave models (fm) = n(v/2L)

Where n is a positive integer and L is the open tube length

Making L the subject of the formula, we have; L = nv/2fm

Now from earlier derivation, we see that v = fλ and in this case, v=fλ

Thus, let's replace v with fλ to het;

L = nλ/2

If we take, n=1, L = (1 x 0.7773)/2 = 0.3887m

Now, when the air inside the tube has warmed, it will have a new length to eliminate beats and give same frequency of 440Hz.

So let's call this new length L1;

So L1 = v'/2(f2) = 346/(2x440) = 346/880 = 0.3932m

So the length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork will be;

ΔL = L1 - L = 0.3932 - 0.3887 = 0.0045m

c. convergent boundary.

explanation: